The function \(g\)  appearing in the representation
(REF ) is identical to the derivative
\(f^{\prime }\)  if \(f\)  is sufficiently smooth.
The representation (REF ) implies, by the
fundamental theorem of calculus (see, e.g., [1]}), that \(f^{\prime }\)  exists and is equal to \(g\)  almost
everywhere (w.r.t the Lebesgue measure) on \([0, 1]\) .
Hence, if \(f \in \mathcal {F}_{\infty -\text{mars}}^1\)  is smooth in the sense that \(f^{\prime }\)  exists everywhere and continuous on \([0, 1]\) , then \(g\)  is identical to \(f^{\prime }\)  and \(V_{\mathrm {mars}}(f) = V(f^{\prime })\) .
For these reasons, somewhat loosely, we can describe the estimator \(\hat{f}_{n, V}^1\)  as
\(\hat{f}_{n, V}^1 \in \mathop {\rm argmin}_{f} \bigg \lbrace \sum _{i=1}^n \big (y_i - f(x^{(i)})\big )^2 : V(f^{\prime }) \le V \bigg \rbrace .\) 
